RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.5

The RD Sharma Class 9 Solutions for the chapter “Circle” Exercise 16.5 are given here. In this exercise, students will learn about cyclic quadrilaterals. A quadrilateral is a cyclic quadrilateral if all its vertices lie on a circle. This exercise includes several questions to help the students learn more about circles. Students are advised to click on the link below and get the PDF to practise more on the topic – Circle. RD Sharma Solutions Class 9 Chapter 16 help students in their endeavour to build a better understanding of circles.

RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.5

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Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 16 Circles Exercise 16.5 Page number 16.83

Question 1: In figure, ΔABC is an equilateral triangle. Find m∠BEC.

Solution:

ΔABC is an equilateral triangle. (Given)

Each angle of an equilateral triangle is 60 degrees.

In quadrilateral ABEC:

∠BAC + ∠BEC = 180^o (Opposite angles of quadrilateral)

60^o + ∠BEC = 180 ^o

∠BEC = 180 ^o – 60 ^o

∠BEC = 120 ^o

Question 2: In figure, Δ PQR is an isosceles triangle with PQ = PR and m∠PQR=35°. Find m∠QSR and m∠QTR.

Solution:

Given: ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°

In ΔPQR:

∠PQR = ∠PRQ = 35^o (Angle opposite to equal sides)

Again, by angle sum property

∠P + ∠Q + ∠R = 180 ^o

∠P + 35 ^o + 35 ^o = 180 ^o

∠P + 70 ^o = 180 ^o

∠P = 180 ^o – 70 ^o

∠P = 110 ^o

Now, in quadrilateral SQTR,

∠QSR + ∠QTR = 180 ^o (Opposite angles of quadrilateral)

110 ^o + ∠QTR = 180 ^o

∠QTR = 70 ^o

Question 3: In figure, O is the centre of the circle. If ∠BOD = 160^o, find the values of x and y.

Solution:

From figure: ∠BOD = 160 ^o

By degree measure theorem: ∠BOD = 2 ∠BCD

160 ^o = 2x

or x = 80 ^o

Now, in quadrilateral ABCD,

∠BAD + ∠BCD = 180 ^o (Opposite angles of Cyclic quadrilateral)

y + x = 180 ^o

Putting value of x,

y + 80 ^o = 180 ^o

y = 100 ^o

Answer: x = 80 ^o and y = 100 ^o.

Question 4: In figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100^o and ∠ABD = 70^o, find ∠ADB.

Solution:

From figure:

In quadrilateral ABCD,

∠DCB + ∠BAD = 180^o (Opposite angles of Cyclic quadrilateral)

100 ^o + ∠BAD = 180^o

∠BAD = 80⁰

In Δ BAD:

By angle sum property: ∠ADB + ∠DAB + ∠ABD = 180 ^o

∠ADB + 80^o + 70 ^o = 180 ^o

∠ADB = 30^o

Question 5: If ABCD is a cyclic quadrilateral in which AD||BC (figure). Prove that ∠B = ∠C.

Solution:

Given: ABCD is a cyclic quadrilateral with AD ‖ BC

=> ∠A + ∠C = 180^o ………(1)

and ∠A + ∠B = 180^o ………(2)

Form (1) and (2), we have

∠B = ∠C

Hence proved.

Question 6: In figure, O is the centre of the circle. Find ∠CBD.

Solution:

Given: ∠BOC = 100^o

By degree measure theorem: ∠AOC = 2 ∠APC

100 ^o = 2 ∠APC

or ∠APC = 50 ^o

Again,

∠APC + ∠ABC = 180 ^o (Opposite angles of a cyclic quadrilateral)

50^o + ∠ABC = 180 ^o

or ∠ABC = 130 ^o

Now, ∠ABC + ∠CBD = 180 ^o (Linear pair)

130^o + ∠CBD = 180 ^o

or ∠CBD = 50 ^o

Question 7: In figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 50⁰, find ∠AOC.

Solution:

Given: ∠OBD = 50⁰

Here, AB and CD are the diameters of the circles with centre O.

∠DBC = 90⁰ ….(i)

Also, ∠DBC = 50⁰ + ∠OBC

90⁰ = 50⁰ + ∠OBC

or ∠OBC = 40⁰

Again, By degree measure theorem: ∠AOC = 2 ∠ABC

∠AOC = 2∠OBC = 2 x 40⁰ = 80⁰

Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(∠CAB) = 30⁰. Find m(∠ACB) and m(∠ABC).

Solution:

Given: m(∠CAB)= 30⁰

To Find: m(∠ACB) and m(∠ABC).

Now,

∠ACB = 90⁰ (Angle in semi-circle)

Now,

In △ABC, by angle sum property: ∠CAB + ∠ACB + ∠ABC = 180⁰

30⁰ + 90⁰ + ∠ABC = 180⁰

∠ABC = 60⁰

Answer: ∠ACB = 90⁰ and ∠ABC = 60⁰

Question 9: In a cyclic quadrilateral ABCD if AB||CD and ∠B = 70^o , find the remaining angles.

Solution:

A cyclic quadrilateral ABCD with AB||CD and ∠B = 70^o.

∠B + ∠C = 180^o (Co-interior angle)

70⁰ + ∠C = 180⁰

∠C = 110⁰

And,

=> ∠B + ∠D = 180⁰ (Opposite angles of Cyclic quadrilateral)

70⁰ + ∠D = 180⁰

∠D = 110⁰

Again, ∠A + ∠C = 180⁰ (Opposite angles of cyclic quadrilateral)

∠A + 110⁰ = 180⁰

∠A = 70⁰

Answer: ∠A = 70⁰ , ∠C = 110⁰ and ∠D = 110⁰

Question 10: In a cyclic quadrilateral ABCD, if m ∠A = 3(m∠C). Find m ∠A.
Solution:

∠A + ∠C = 180^o …..(1)

Since m ∠A = 3(m∠C) (given)

=> ∠A = 3∠C …(2)

Equation (1) => 3∠C + ∠C = 180 ^o

or 4∠C = 180^o

or ∠C = 45^o

From equation (2)

∠A = 3 x 45^o = 135^o

Question 11: In figure, O is the centre of the circle ∠DAB = 50°. Calculate the values of x and y.

Solution:

Given : ∠DAB = 50^o

By degree measure theorem: ∠BOD = 2 ∠BAD

so, x = 2( 50⁰) = 100⁰

Since, ABCD is a cyclic quadrilateral, we have

∠A + ∠C = 180⁰

50⁰ + y = 180⁰

y = 130⁰

RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.5

RD Sharma Solutions Class 9 Maths Chapter 16 Circles Exercise 16.5 is based on the following topics and subtopics:

• Cyclic Quadrilateral
• Important Results:

-The opposite angles of a Cyclic Quadrilateral are supplementary.

– The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

– A cyclic trapezium is isosceles and its diagonals are equal.

– If two opposite sides of a cyclic quadrilateral are equal, then the other two sides are parallel.